Question

Given the following atomic weights, that the density of acetic acid is 1.05, and that 7.0 g of isoamyl alcohol, 6.6 mL of acetic acid, and 0.5 mL of sulfuric acid make 3.2 g of isoamyl acetate, what is the % yield of isoamyl acetate from its limiting reagent? Give at least two significant figures.

Answer 1

Complete reaction is:

Volume of acetic acid = 6.6 mL

density of acetic acid = 1.05 g/mL

Mass of acetic acid = volume * density = 6.6 mL * 1.05 g/mL = 6.93 g

Molar mass of acetic acid =60.05 g/mol

Moles of acetic acid = Mass/ Molar mass = 6.93 g /(60.05 g/mol) = 0.115 mol

Mass of isoamyl alcohol = 7 g

Molar mass of isoamyl alcohol = 88.148 g/mol

Moles of isoamyl alcohol = 7 g/ (88.148 g/mol)= 0.0794 mol

H₂SO₄ is just a catalyst

From reaction, 1 mol of isoamyl alcohol reacts with 1 mol of acetic acid

Thus, 0.115 mol of isoamyl alcohol reacts with 0.115 mol of acetic acid

But we have 0.0794 mol of acetic acid. Thus, acetic acid is a limiting reagent.

1 mol of acetic acid produces 1 mol of isoamyl acetate

So, 0.0794 mol of acetic acid produces 0.0794 mol of isoamyl acetate

Molar mass of isoamyl acetate = 130.19 g/mol

Mass of isoamyl acetate produced = Moles * Molar mass = 0.0794 mol * 130.19 g/mol = 10.3 g

Thus, theoretical yield = 10.3 g

actual yield = 3.2 g

Hence percent yield = (Actual yield / theoretical yield )*100%

= (3.2g/ 10.3 g)*100% = 30.95 % = 31 % ( in 2 significant figures)